Why is $x^2\sin(1/x)$ not strictly differentiable? I am on Wikipedia reading on strict differentiability and I don't particularly understand the example proving a function that is differentiable does not have to be strictly differentiable, where $f(x)=x^2 \sin(1/x)$, $f(0)=0$. How can we show this is differentiable directly using difference quotients. Also then how can we show it's not strictly differentiable using our difference quotients?

The simplest setting in which strict differentiability can be considered, is that of a real-valued function defined on an interval $I$ of the real line. The function $f: I \rightarrow \mathbf{R}$ is said strictly differentiable in a point $a \in I$ if
$$
\lim _{(x, y) \rightarrow(a, a)} \frac{f(x)-f(y)}{x-y}
$$
exists, where $(x, y) \rightarrow(a, a)$ is to be considered as limit in $\mathbf{R}^{2}$, and of course requiring $x \neq y$.
A strictly differentiable function is obviously differentiable, but the converse is wrong, as can be seen from the counter-example $f(x)=x^{2}$ sin $\frac{1}{x}, f(0)=0, x_{n}=\frac{1}{\left(n+\frac{1}{2}\right) \pi}, y_{n}=x_{n+1}$.
One has however the equivalence of strict differentiability on an interval $I$, and being of differentiability class $C^{1}(I)$.
(Transcribed from Wikipedia screenshot)

so far I have;
To show $f(x)$ is differentiable, I would assume I'd use the statement that if $x^2$ is differentiable and $\sin(1/x)$ is differentiable (which can be shown by the difference quotients, then $x^2 \sin(1/x)$ is also differentiable?
 A: Recall negation of function limit definition.
$$
\text{if } \exists p_k, q_k\in\mathbb{R^n}:
\lim_{k\rightarrow \infty} p_k, q_k=x_0:
\lim_{{k\rightarrow \infty}} g(p_k) \neq \lim_{{k\rightarrow \infty}} g(q_k)
\Rightarrow \text{no limit}
$$
Let us use this fact to prove that the desired limit doesn't exist.
$$
\begin{multline}
\exists
p_k=\left(\begin{smallmatrix}x_{2k}\\y_{2k}\end{smallmatrix}\right),
q_k=\left(\begin{smallmatrix}x_{2k+1}\\y_{2k+1}\end{smallmatrix}\right)
\in \mathbb{R}^2:
\lim_{k\rightarrow \infty} p_k, q_k=
\left(\begin{smallmatrix}0\\0\end{smallmatrix}\right):\\
\frac{2}{\pi}=\lim_{{k\rightarrow \infty}} \frac{f(x_{2k})-f(y_{2k})}{x_{2k}-y_{2k}} \neq \lim_{{k\rightarrow \infty}}  \frac{f(x_{2k+1})-f(y_{2k+1})}{x_{2k+1}-y_{2k+1}}=-\frac{2}{\pi}
\end{multline}
$$
where $f(z) = z^2\sin\frac{1}{z}$, $x_n=\frac{1}{(n+1/2)pi}$ and $y_n=x_{n+1}$.
$$
\begin{aligned}
\lim_{x \rightarrow 0\\y \rightarrow 0}\cfrac{x^2\sin\frac{1}{x} - y^2\sin\frac{1}{y}}{x-y}
&=\lim_{n \rightarrow \infty}\cfrac{x_n^2\sin\frac{1}{x_n} - y_n^2\sin\frac{1}{y_n}}{x_n-y_n}\\
&=
\lim_{n \rightarrow \infty}\cfrac{\left(\frac{1}{(n+1/2)\pi}\right)^2\sin(n+1/2)\pi - \left(\frac{1}{(n+3/2)\pi}\right)^2\sin(n+3/2)\pi}{\frac{1}{(n+1/2)\pi}- \frac{1}{(n+3/2)\pi}}\\
&=
\lim_{n \rightarrow \infty}\cfrac{\left(\frac{1}{(n+1/2)\pi}\right)^2(-1)^n - \left(\frac{1}{(n+3/2)\pi}\right)^2(-1)^{n+1}}{\frac{1}{(n+1/2)(n+3/2)\pi}}\\
&=
\lim_{n \rightarrow \infty}\left[\left(\frac{1}{(n+1/2)\pi}\right)^2 + \left(\frac{1}{(n+3/2)\pi}\right)^2\right](-1)^{n}(n+1/2)(n+3/2)\pi\\
&=
\lim_{n \rightarrow \infty}\frac{2n^2+4n+5/2}{(n+1/2)^2(n+3/2)^2\pi^2}(-1)^{n}(n+1/2)(n+3/2)\pi\\
&=
\lim_{n \rightarrow \infty}\frac{n^2+2n+5/4}{(n+1/2)(n+3/2)}\frac{(-2)^{n}}{\pi}\\
\end{aligned}
$$
It is clear to see, that
$$
\begin{aligned}
n = 2k &\rightarrow p_k \rightarrow\lim_{n \rightarrow \infty} \cfrac{\sim 2n^2}{\sim \pi n^2} = \frac{2}{\pi}\\
n = 2k+1 &\rightarrow q_k \rightarrow\lim_{n \rightarrow \infty} -\cfrac{\sim 2n^2}{\sim \pi n^2} = -\frac{2}{\pi}
\end{aligned}
$$
To sum up, $f$ is not strictly differentiable.
A: More general result: Suppose $f$ is differentiable on an open interval $I,$ and $f'$ is discontinuous at some $x_0\in I.$ Then $f$ is not strictly differentiable on $I.$
Proof: Let $a_n$ be a sequence of distinct points converging to $x_0.$  By the MVT there exist $b_n$ between $x_0$ and $a_n$ such that
$$\frac{f(a_n)-f(x_0)}{a_n-x_0}=f'(b_n).$$
As $n\to \infty,$ we see $f'(b_n) \to f'(x_0).$ On the other hand, because $f'$ is discontinuous at $x_0,$ there exists a sequence $c_n\to x_0$ such that $f'(c_n)$ does not converge t0 $f'(x_0).$
Now find sequences $s_n,t_n\to x_0$ such that
$$\left | \frac{f(s_n)-f(t_n)}{s_n-t_n}-f'(b_n) \right |<1/n$$
and sequences $u_n,v_n\to x_0$ such that
$$\left | \frac{f(u_n)-f(v_n)}{u_n-v_n}-f'(c_n) \right |<1/n.$$
Both the sequences $(s_n,t_n),(u_n,v_n)$ converge to $(x_0,x_0).$ Consider the sequence $(x_n,y_n)$ obtained by putting these two sequences together. I.e., $(x_n,y_n)$ is the sequence
$$(s_1,t_1),(u_1,v_1),(s_2,t_2),(u_2,v_2), \dots $$ 
Then $(x_n,y_n)\to (x_0,x_0)$ and the corresponding difference quotients converge to $f'(x_0)$ through odd values of $n,$ but fail to converge to $f'(x_0)$ through even values of $n.$
Therefore $f$ is not strictly differentiable on $I.$
In our specific problem $f'$ is not continuous at $0,$ so this $f$ is not stricly differentiable on $\mathbb R$ by the general result.
